Determining Empirical and Molecular Formulas

Empirical Formula: a GENERAL formula of a compound in its most simplified whole # ratio of atoms.

Molecular Formula: A SPECIFIC non-simplified formula of a compound.

Empirical Formulas:

Empirical formulas may be molecular formulas themselves or they may be a building block or unit of a larger molecular formula.

For example, C5H12 cannot be simplified any further because 5 and 12 are not divisible by any common number. Therefore, C5H12 is both an empirical and molecular formula.

But, is C2H6 can empirical formula? No, because you can simplify the formula by dividing the subscripts each by a common multiple of 2 to form: CH3­. Therefore, CH­3 is an empirical formula while C2H6 is a molecular formula.

To determine the empirical formula of a compound, you will typically be given the following:

Masses of each element in the formula OR

Percentages of each element in the formula

Use this information and following steps to calculate the empirical formula.

Steps to Determining an Empirical Formula:

If given gram amounts, you are one step ahead. If given percentages, ensure they all add up to 100%.

Convert all percentages to grams using a 1:1 ratio (ie, 20% = 20g) . As long as everything adds up to the same value (100% or 100g) then it the same ratio mathematically. So the % of each atom can substitute for the gram amount.

Divide grams of each element by its atomic mass to determine the # of units of that element exist in the ratio.

Once all subscripts (ratios) for each element have been determined in step 3, simplify the ratios by dividing each value by the smallest value. This will guarantee that you have a “1” as one of your subscripts and that you will have the most simplified formula.

Are your simplified ratios whole #s? If yes, rewrite the equation using your simplified ratios. This is the empirical formula!

If the ratios are not whole #, then multiply that ratio by the smallest number possible so that it will give you a whole number (ie 1.33 x 3 = 4 or 1.5 x 2 = 3). NOTE: you will also have to multiply all the other ratios by the same multiple in order to keep the ratio consistent.

STEP 3: Divide grams of each element by its atomic mass to determine the # of units of that element exist in the ratio.

C: 40%

40 g/12 g = 3.33 units

H: 6.7%

6.7g/1 g = 6.7 units

O: 53.3%

53.3 g/16 g = 3.33 units

STEP 4:

C: 40%

40 g/12 g= 3.33 mols / 3.33 = 1

1 C

H: 6.7%

6.7g/1 g= 6.7 mols /3.33 = 2

2 H

O: 53.3%

53.3 g/16 g= 3.33 mols/3.33 = 1

1 O

STEP 5: Since the simplified ratios are whole #s (1, 2, 1) , rewrite the equation using your simplified ratios. This is the empirical formula!

CH­2O

Molecular Formulas:

You can use the empirical formula to find the molecular formula. Recall that the molecular formula is simply a MULTIPLE of the empirical formula, meaning it has the “X” times the number of all atoms in the empirical formula, and therefore, “X” times the mass of the empirical formula.

You will be using the ratio of the empirical formula’s mass, called the empirical mass, and the molecular formula’s mass given, the molecular mass or weight, to find out just how many times bigger is the molecular formula to the empirical formula?

Steps to Determining an Molecular Formula:

Find the mass of the empirical formula. This is done just as you would find formula weight.

Divide the molecular mass by the empirical mass to find your multiple.

Multiply the empirical formula’s subscripts by the multiple.

Example: What is the molecular formula of a compound which contains 40% carbon, 6.7% hydrogen, and 53.3% oxygen and has a molecular weight is 180 grams? [Ans: C6H12O6]

STEP 1: Determine the empirical formula if you have not already done so. Then calculate its mass.

About ChemistryBytes

ChemistryBytes was started as a simple and direct way to help students digest Chemistry concepts in "byte" sized portions. ChemBytes, as it is known, features write ups on various Chemistry concepts, as well as videos that make following along a fun and easy task.